Hydraulic force question

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Umm, hydraulic pistons have two important characteristics: force and stroke. Two identical cylinders with equal pressures placed in parallel double force, the same two in series double stroke. Sort of the same way identical batteries in parallel double current, the same in series double voltage. Likewise adding links to a chain makes it longer, not stronger. The discussion was about stationary brake calipers with opposed pistons--effectively hydraulic cylinders in series. So you would get a potential increase in stroke, but no increase in force. But since a fixed brake caliper needs no more total stroke than a floating caliper it's more comparable to replacing one long cylinder with two short ones stacked on top of each other. Someone did mention earlier in the discussion a caliper with parallel cylinders. He also mentioned that the cylinders are reduced in size--that keeps down the force on the pads and reduces the amount of fluid the master cylinder would need to move to apply the brakes. But that's a different arrangement. You mentioned two bottle jacks, but did not say if they were in series or parallel. So, yeah, under the right circumstances two, ten, or a hundred one ton bottle jacks push no harder than a single one ton jack. The formula is more correctly written as Force = piston cross sectional area X pressure - friction. Pad psi = force / pad cross sectional area. You'll note that pads follow the same rule: two pads in series have the same psi as one pad with the same area. But you get more friction since friction force = total pad area X pad psi.

Reply to
B.B.
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In other words, if you stacked your two one ton car jacks nose to nose,=20 rather than nose to base, you'd be able to lift 2 tons??

In your ASCII diagram, no matter where in the stack you insert a force=20 sensor, it will register the same value. If this were not the case=20 there'd have to be some additional reaction to ground between the rigid=20 surfaces.

Ned Simmons

Reply to
Ned Simmons
[stacked jacks]

They are in series. Sorry, but if you don't know the difference between serial and parallel, further discussion seems quite useless.

Um, ah! So if you stack two jacks and pump one with 1ton you will have 1 ton pressure between the two jacks. Then when you pump the second jack to 1 ton (disregarding it has already a "pressure" of one ton) the force will not increase, but the pressure between will be 2 tons? Get real!

This is my last attempt to explain: First, I ask you to read back a bit and try to understand my posting with the "two opposed spindles C-clamp" example. Thanks.

Now, you have to accept a fact of physics. "actio = reactio". Or, in plain english, "every force has a counter-force (or reacting force)". More verbose: Every force has a reacting force of the same value, but opposed direction and colinear (in line) with that force. Or, the vectorial summ of the forces must be zero. Accept that? OK

Low lets look at the single piston bracket:

----->DDDDDD YADRO

Reply to
Nick Müller

"Bruce L. Bergman" wrote

I can't believe this thread is still going. In your sketch, the vertical _force_ on the rotor is ZERO. That's what you _want_, unless your rotor is on a mighty strong bearing. That's what you _get_, whether you use two "rigid surfaces" and two jacks, or whether you use two "floating surfaces" rigidly connected to each other, and replace one jack with a spacer.

The _horizontal_ force on your rotor is your braking force. It is your friction coefficient times _two_ tons, because there are _two_ sliding interfaces. But there would be two interfaces even if you replace one jack with a spacer and let the caliper float.

Replace one of your "pads" with a (frictionless) _bearing_, and you cut your braking force in half. But the net vertical force _on_ the rotor is still zero, and the net vertical force _through_ the rotor is still one ton.

-- TP

Reply to
tonyp

Did I miss something here or did you mean to respond to different person?

Reply to
tomcas

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